A long spring is stretched by 2 cm, its potential energy is U. If the spring is stretched by 10 cm, the potential energy stored in it will be

The correct option is 25 U.

Step-by-step Explanation:

The elastic potential energy stored in a stretched spring is given by the formula:

$U=\frac{1}{2}k{x}^2$

where:
- $U$ is the potential energy stored in the spring,
- $k$ is the spring constant (stiffness of the spring), and
- $x$ is the displacement or stretch length of the spring.

For a given spring, the spring constant $k$ is constant. Therefore, the potential energy stored in the spring is directly proportional to the square of its extension:

$U\propto x^2$

Let $U_1$ be the initial potential energy when the spring is stretched by $x_1=2\text{cm}$, so $U_1=U$.
Let $U_2$ be the final potential energy when the spring is stretched by $x_2=10\text{cm}$.

We can write the ratio of the potential energies as:

$\frac{U_2}{U_1}={\left(\frac{x_2}{x_1}\right)}^2$

Substitute the given values into the equation:

$\frac{U_2}{U}={\left(\frac{10}{2}\right)}^2$

Simplify the fraction inside the parentheses:

$\frac{U_2}{U}=5^2$

$\frac{U_2}{U}=25$

Solving for $U_2$:

$U_2=25U$

Thus, the potential energy stored in the spring when it is stretched by 10 cm will be 25 U.